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A positive gate voltage attracts electrons towards the oxide interface of the p-type semiconductor and repels mobile electron holes. The uncovered ionized acceptors form a negatively charged depletion region. Increasing the voltage further creates a thin electron-rich semiconductor inversion layer at the surface. The conduction band and valence band both bend downwards towards the positively biased gate, because an electron's electrostatic energy is $-e\psi$.

In an ideal nondegenerate substrate with acceptor density $N_A$, write ${\varphi }_F=(k_{B}T/e)\ln (N_A/n_i)>0$ and surface potential ${\psi }_s$ relative to the neutral bulk. The electron and hole surface densities areThe first change of majority-carrier type occurs at ${\psi }_s={\varphi }_F$, where $n_s=p_s=n_i$. The conventional MOS inversion threshold for appreciable strong inversion isAt threshold the inversion sheet is just developing, while a much thicker depletion region lies between it and the neutral p-type bulk. Inversion does not require the conduction band to cross the Fermi level: the surface can remain nondegenerate.

The metal and semiconductor Fermi levels need not be equal. The ideal oxide blocks particle exchange, so the gate and substrate can maintain different electrochemical potentials. In a biased MOSFET, the gate-to-substrate Fermi-level difference is ${\tilde{\mu }}_g-{\tilde{\mu }}_s=-e{V}_g$, when the applied voltage is defined in the electron-reservoir convention. Each region can separately be in internal equilibrium. If an external zero-voltage connection instead equilibrates the reservoirs, their Fermi levels coincide, but different work functions can still cause band bending. Here the chemical potential shown in a band diagram is the electron electrochemical potential; the purely chemical contribution alone is not the equilibrium criterion for charged particles.

Equilibrium junction. Put the p-type side on the left and n-type side on the right. When they are joined, electrons diffuse from n to p and electron holes diffuse from p to n. This leaves negative ionized acceptors on the p-side and positive ionized donors on the n-side of the depletion region. The resulting built-in electric field points from n to p, opposing further diffusion. At equilibrium there is one constant Fermi level, and drift and diffusion currents cancel for each carrier species.

In the nondegenerate, fully ionized limit, the neutral-region carrier densities areThe conduction band and valence band lie higher on the p-side and bend down towards the n-side. The gap between them stays fixed for a homojunction. If $\psi$ increases towards the n-side, both edges shift by $-e\psi$, whileThus $n$ rises and $p$ falls continuously across the junction; neither mobile population is an exact step function. Their depletion exposes fixed dopant charge density, rather than an absence of every kind of charge. The built-in voltage is

Light emission. Forward bias lowers the barrier and injects minority carriers across the p-n junction. In a light-emitting diode, electron-hole recombination converts some of their energy into photons. A direct band gap permits efficient radiative transitions with nearly conserved crystal momentum; nonradiative channels compete. Near-edge emission has $\hslash \omega \simeq E_g$ and $\lambda \simeq hc/E_g$. Material composition and confinement select the emission colour. Reverse-bias saturation is not the ordinary LED operating mode.

Reverse current. Take $V=V_R>0$ and $I_R>0$ to mean reverse voltage and reverse current magnitudes. The Shockley diode equation in this convention isThe bias-independent $I_S$ term is the supply of thermally excited minority carriers from the neutral regions, swept across the junction by its electric field. The opposing injection current is reduced by the extra barrier energy $e{V}_R$ and is $I_S{e}^{-e{V}_R/(k_{B}T)}$. Their equality gives zero net current at zero bias; for $e{V}_R\gg k_{B}T$, the reverse current approaches $I_S$ until leakage or breakdown invalidates this ideal law. The collecting field need not create the minority carriers; their thermal supply limits the diffusion-dominated current.

For an ideal long junction of area $\mathcal{A}$, steady diffusion with recombination gives the diode minority-carrier diffusion currentIt depends on doping, geometry, carrier diffusion coefficients and lifetimes, and especially the intrinsic carrier concentration. Depletion-region generation current and surface leakage can add nonideal contributions to the measured reverse current.

Band-gap estimate from the data. Read a nearly flat region well above the small-bias knee and below the high-bias upturn, rather than estimating $I_S$ in the breakdown region. For a diffusion-dominated junction with nearly temperature-independent gap and effective masses,so $I_S=C(T)T^3{e}^{-E_g/(k_{B}T)}$, where $C(T)$ includes transport and doping factors. Extracting a semiconductor band gap from diode saturation current amounts to measuring the slope of $\ln [I_S/(C(T)T^3)]$ against $1/T$: it is $-E_g/k_B$. If the prefactor is taken approximately constant over a short interval,Approximate readings from the original PDF plateaus areUsing the two warmer curves and $k_B=8.6173\times 10^{-5}eV/K$ givesKeeping the $T^3$ factor but taking the residual transport factor constant changes this estimate to approximately $0.57\mathrm{eV}$. The justified graphical estimate is thereforeThe colder curve is a genuine limitation of the supplied data: the $0$ to $-45{}^{\circ }\mathrm{C}$ pair gives only about $0.32\mathrm{eV}$ with a constant prefactor, and the endpoint pair gives about $0.42\mathrm{eV}$. Thus all three cannot support one precise diffusion-only activation slope. A depletion-region generation current, approximately proportional to $n_i$ rather than $n_i^2$, can have a roughly half-gap activation energy; other leakage and temperature-dependent transport can also alter the slope. The PDF alone does not identify their relative contributions. A reliable precision gap measurement would use more temperatures and verify the transport regime instead of forcing one fit through these three readings.

At equilibrium, majority electrons undergo diffusion from n to p, while majority electron holes diffuse from p to n. Their departure exposes positive donors on the n side and negative acceptors on the p side. The resulting electric field points from n to p and opposes further diffusion. Equilibrium is reached when the chemical potential is constant across the junction. The built-in potential is this electrostatic adjustment, not a discontinuity of the material's band gap.

Since electronic band energies shift by $-eV(x)$, both the conduction band and valence band undergo band bending down from p to n by $e{V}_b$ at zero applied bias. Their separation remains the same gap. In the nondegenerate thermal equilibrium limit the built-in voltage can also be expressed using neutral-edge densities as $e{V}_b=k_{B}T\log (n_n/n_p)=k_{B}T\log (n_n{p}_p/n_i^2)$.

Take the n-contact potential and its electrochemical potential as reference. Applying $V_{\mathrm{ext}}$ to the p side changes the total depletion-layer drop toThus $V_{\mathrm{ext}}<0$ is reverse bias: it increases the barrier and widens the depletion region. The bands have horizontal tangents at the neutral edges in this approximation and a smooth cubic bend within the linearly graded layer.

A biased junction is not in global thermal equilibrium. Its contact electron electrochemical potentials obeyUnder reverse bias the p-contact level is higher by $e|V_{\mathrm{ext}}|$. A single horizontal equilibrium chemical potential cannot describe this state; electron and hole quasi-Fermi levels describe the nonequilibrium populations. The sketch marks the contact levels. Detailed spatial quasi-Fermi profiles require transport and recombination information beyond the electrostatic doping profile.

A concrete example is samarium hexaboride, ${\mathrm{SmB}}_6$: narrow, nearly localized $4f$-derived states mix with a much more dispersive $5d$-derived conduction band. Such band hybridization is the low-energy mechanism behind a bulk Kondo insulator gap. The two-band Hamiltonian is an effective illustration; correlations help set its narrow-band energy and renormalized mixing.^[ref]

Count wavevectors in a large volume with periodic boundary conditions. One spin-degenerate spherical band suppliesFor the conduction band, $k=\sqrt{2m_e^{\ast }(ϵ-{ϵ}_c)}/\hslash$ and $dk/dϵ=m_e^{\ast }/({\hslash }^{2}k)$. Transforming the state count to energy givesFor the valence band use $|dk/dϵ|=m_h^{\ast }/({\hslash }^{2}k)$ and $k=\sqrt{2m_h^{\ast }({ϵ}_v-ϵ)}/\hslash$, obtainingEach density of states is zero beyond its indicated band edge. The factor two counts spin; an extra valley degeneracy was not assumed in these stipulated isotropic bands.

In the nondegenerate conduction band, the Fermi-Dirac distribution reduces to its Boltzmann distribution tail. ThusHere the substitution $x=(ϵ-{ϵ}_c)/(k_{B}T)$ leaves ${\int }_0^{\infty }{x}^{1/2}{e}^{-x}dx=\sqrt{\pi }/2$. For electron holes use the empty-state probability $1-f(ϵ)\simeq e^{(ϵ-\mu )/(k_{B}T)}$, not the conduction-band Boltzmann factor. This givesThe chemical potential cancels in the product. With $E_g={ϵ}_c-{ϵ}_v$,This semiconductor mass-action law holds in thermal equilibrium for nondegenerate carriers, including a doped semiconductor. Only in an intrinsic semiconductor must $n=p=n_i$. The factors $N_c,N_v$ are the effective density of states; the factor $1/2$ in the final expression follows from the two spin factors already included in (a).

A tunnel diode is a p-n junction with sufficiently heavy semiconductor doping on both sides to make a thin depletion region and permit substantial interband tunnelling. The neutral regions are degenerate semiconductors. Its distinctive forward current-voltage characteristic contains a peak followed by a valley: over part of the curve, increasing the applied electric potential difference decreases the electric current. The Esaki tunnel-diode experiment made this behavior experimentally concrete.

The useful quantity is the local slope, not the ratio of total voltage to total electric current. With conventional electric current $I$ entering the positive-voltage terminal, defineNegative differential resistance means $g_d<0$, hence $r_d<0$ where defined. A forward operating point can still have $V>0$, $I>0$ and absorbed power $VI>0$. A small voltage increase then gives a negative current increment; this does not make the total electric current negative or turn the device into an unlimited source of energy. The external supply sets the operating point and supplies the energy available for an electronic amplifier.

An original idealized construction explains the shape. Put the n-type semiconductor on the left and the p-type semiconductor on the right. Take $q>0$ as the elementary charge and $V=V_p-V_n$. Increasing forward bias lowers the right-hand electron energy levels relative to the left. Denote the two reservoir electrochemical potentials by ${\mu }_n,{\mu }_p$, so that ${\mu }_n-{\mu }_p=qV$. At zero temperature, occupied left-hand conduction band states extend up to ${\mu }_n$; empty right-hand valence band states start at ${\mu }_p$. For transfer conserving electron energy, the available interval isIt is nonempty only when its upper endpoint exceeds its lower endpoint. Both an occupied initial quantum state and an empty final quantum state must exist at the same energy. The Pauli exclusion principle prevents transfer into an already occupied final quantum state.

For the original diagram, choose arbitrary energy units with $E_{c,n}=-0.5$, ${\mu }_n=0$, ${\mu }_p=-u$ and $E_{v,p}=0.5-u$, where $u$ is $qV$ in those units. The available interval has widthThus $L(0.25)=0.25$, $L(0.50)=0.50$, $L(0.70)=0.30$, and $L(1.15)=0$. Initially forward bias opens a transfer window; further forward bias closes it by moving the right-hand valence band edge below the left-hand conduction band edge. This is a state-counting illustration, not a formula for a measured electric current.

For reverse bias, the transfer direction changes: occupied right-hand valence band states can feed empty left-hand conduction band states. In the same construction, $u=-0.70$ opens the interval $0<E<0.70$ for this reverse transfer. A narrow depletion region allows a large reverse quantum tunnelling electric current at small electric potential difference, whereas the ordinary Shockley diode equation predicts only a small reverse saturation electric current before diode reverse breakdown. At zero biasing, equal reservoir occupations give zero net electric current; quantum tunnelling does not require a net equilibrium flow.

A finite-temperature refinement replaces sharp occupation cutoffs with the Fermi-Dirac distribution. An illustrative weak-transfer expression isHere $\mathcal{W}\ge 0$ describes allowed initial/final quantum states and their coupling, and $f_n,f_p$ are Fermi-Dirac distributions with the respective electrochemical potentials. The occupation difference follows by subtracting the reverse transfer factor from the forward one: $f_n(1-f_p)-f_p(1-f_n)=f_n-f_p$. An overlap interval alone does not determine $\mathcal{W}$: transmission probability, density of states, crystal momentum selection and phonons can matter. A complete germanium calculation is more demanding than multiplying two constant density of states factors. At larger forward bias, ordinary carrier injection and diffusion supply a rising current branch even after direct band overlap is lost.

A type-I band alignment confines electrons and electron holes together: its inner conduction band edge is lower, its valence band edge higher.^[ref]

In the chosen band offset example, the electron barrier is $0.6$, while the electron hole barrier is $0.4$. A hole's excitation energy is measured downward from a valence band maximum: reversing that convention would incorrectly make the two barriers appear to repel opposite carriers. An electrostatic shift $-q\varphi$ moves both electron energy band edges together; changing the band gap also changes their separation. Thus material composition supplies a control beyond the electric potential $\varphi$ alone.

A Wannier-Mott exciton has an internal wave function extending over many unit cells. Its simplest model assumes a three-dimensional bulk crystal, approximately parabolic valence band and conduction band, positive isotropic scalar effective masses $m_e^{\ast },m_h^{\ast }$, a constant scalar relative permittivity ${ϵ}_r$, and a dilute pair population. Replace microscopic crystal forces by those effective masses and dielectric screening, and neglect exchange corrections and coupling to phonons. These assumptions must hold on the spatial and energy scales of the particular exciton.

With relative coordinate $\mathbf{r}={\mathbf{r}}_e-{\mathbf{r}}_h$, the internal Hamiltonian operator is the three-dimensional hydrogen atom problem with a reduced mass and a screened Coulomb potential:Here $e$ is the elementary charge, ${ϵ}_0$ the vacuum permittivity, and $\hslash$ the reduced Planck constant. The internal energy eigenvalues, radius scale, and exciton binding energies areThe principal quantum number is $n=1,2,\dots$. This reduction and these scales are derived in Excitons in Semiconductor Quantum Dots by Matjaž Humar (§3, printed pp. 5–6, equations (13)–(26)).

To check the scales by dimensional analysis, write $C=e^2/(4\pi {ϵ}_0{ϵ}_r)$, which has units of energy times length. Then $a_B^{\ast }={\hslash }^2/(\mu C)$ is a length and $R^{\ast }=\mu C^2/(2{\hslash }^2)$ is an energy. Rescaling $r=a_B^{\ast }\rho$ makes $H_{\mathrm{rel}}/R^{\ast }=-{\nabla }_{\rho }^2-2/\rho$, the dimensionless hydrogen atom operator. The Bohr radius $a_B^{\ast }$ is a wave function scale, not the radius of a classical orbit; for this model's lowest internal bound state, the mean separation is $3a_B^{\ast }/2$.

For the diagrams' hypothetical inputs $m_e^{\ast }=0.10m_0$, $m_h^{\ast }=0.40m_0$, ${ϵ}_r=10$ and $E_g=1.500\mathrm{eV}$, where $m_0$ is the free electron mass, independent substitution gives $\mu =0.080m_0$, $a_B^{\ast }\simeq 6.615\mathrm{nm}$ and $R^{\ast }\simeq 10.885\mathrm{meV}$. Here electronvolts express energy. The exciton binding energies for $n=1,2,3$ are approximately $10.885$, $2.721$ and $1.209\mathrm{meV}$. These are model calculations, not measured material parameters. The large-radius approximation additionally requires $a_B^{\ast }$ to be much larger than the crystal's unit cell length scale. In a thin layer, with anisotropic effective masses, or with spatially varying dielectric screening, this three-dimensional formula is not generally valid.